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# Making Light - PowerPoint PPT Presentation

Making Light. We can also make light by exciting atoms. From experiment, we see that different atoms emit different light. But each type of atom emits a very specific set of wavelengths called a spectrum .

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## PowerPoint Slideshow about ' Making Light' - xenon

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Presentation Transcript

We can also make light by exciting atoms.

From experiment, we see that different atoms emit different light. But each type of atom emits a very specific set of wavelengths called a spectrum.

The hydrogen atoms emit three visible wavelengths: one in the red, one in the blue-green, and one in the violet.

We need a model of the atom that will explain why atoms emit only certain wavelengths.

First of all, what is the size of a typical atom? Let’s take water (although that is a molecule, we know a lot about water: its mass density: 1 gm / 1 cc, it is H2O so it has 18 grams/mole, and we know Avagadro’s number = 6.02 x 1023 molecules/mole.

(1 cc / 1 gm) * (18 gms / mole) * (1 mole / 6.02 x 1023 molecules)

= 18 x 10-6 m3 / 6 x 1023 molecules

= 3 x 10-29 m3 = 30 x 10-30 m3 .

Therefore, the size is about (30 x 10-30 m3)1/3

= 3 x 10-10 m. Thus the size of an atom should roughly be about 0.1 nm .

Now that we know the size of an atom, how much mass does the atom have?

From the mass spectrograph, we know that the mass of an atom comes in integer values of 1 amu = 1.66 x 10-27 kg. (In fact, this is important in getting Avagadro’s number!)

Now that we know the size and mass, what parts does an atom consist of?

We know that the atom has electrons of very small mass (me = 9.1 x 10-31 kg), about 2000 times smaller than one amu and a negative charge of -1.6 x 10-19 Coul.

We also know that the atom is neutral, so the part of the atom that is not the electrons must have essentially all the mass and a positive charge to cancel that of the electrons.

But what is the structure of these electrons and this other part of the atom?

Two possibilities come to mind:

• The planetary model, where the very light electron orbits the heavy central nucleus.

• The plum pudding model, where the very light and small electrons are embedded (like plums) in the much more massive pudding of the rest of the atom.

The Planetary Model:

If the light electron does go around the central, heavy nucleus, then the electron is accelerating (changing the direction of its velocity). But an accelerating electron should emit electromagnetic radiation (its electric field is wobbling).

• If the electron is emitting E&M radiation, it is emitting energy.

• If the electron is emitting energy, it should then fall closer to the nucleus.

• The process should continue until the electron falls into the nucleus and we have the plum pudding model

• In addition, the frequency of the E&M radiation (light) emitted by the accelerating (orbiting) electron should continuously vary in frequency as the frequency of the electron continuously varies as it spirals into the nucleus. This does not agree with the experimental results: the spectrum of hydrogen.

The plum pudding model has no such problem with accelerating electrons, since the electrons are just sitting like plums in the pudding.

To test the plum pudding model, Rutherford decided to shoot alpha particles

(mass = 4 amu’s; charge = +2e; moving very fast)

at a thin gold foil and see what happens to the alpha particles.

(gold can be made very thin - only several atoms thick; thus there should be very few multiple scatterings)

If the plum pudding model was correct, then the alphas should pretty much go straight through - like shooting a cannon ball at a piece of tissue paper. The positive charge of the atom is supposed to be spread out, so by symmetry it should have little effect. The electrons are so light that they should deflect the massive alpha very little.

Results:

• Most of the alphas did indeed go straight through the foil.

• However, a few were deflected at significant angles.

• A very few even bounced back!

(Once in a while a cannot ball bounced back off the tissue paper!

The results of the scattering were consistent with the alphas scattering off a tiny positive massive nucleus rather than the diffuse positive pudding.

The results indicated that the positive charge and heavy mass were located in a nucleus on the order of 10-14 m (recall the atom size is on the order of 10-10 m).

If the electric repulsion of the gold nucleus is the only force acting on the alpha

(remember both alpha and the nucleus are positively charged)

then the deflection of the alpha can be predicted.

The faster we fire the alpha, the closer the alpha should come to the gold nucleus.

1/2 m v2 = q(kqgold/r)

We will know that we have “hit” the nucleus (and hence know its size) when the scattering differs from that due to the purely electric repulsion. This also means that there must be a “nuclear force”!

Note how small the nucleus is in relation to the atom: the nuclear radius is 10-14 m versus the atomic radius of 10-10 m - a difference in size of 10,000 and a difference in volume of 1012 (a trillion!).

The electron is even smaller. It is so small that we can’t yet say how small, but it is

less than 10-17 meters in radius.

If the mass takes up only 1 trillionth of the space, why can’t I walk right through the wall?

The electric repulsion between the orbiting electrons of the wall and the orbiting electrons of me - and the electric repulsion between the nuclei of the atoms in the wall and the nuclei of my atoms, these repulsions keep me and the wall separate.

The nuclear force does not come into play. We’ll say more about the nuclear force in part V of this course.

We now know that the atom seems to have a very tiny nucleus with the electrons somehow filling out the size of the atom - just what the planetary model of the atom would suggest.

However, we still have the problem of how the electrons stay in those orbits, and how the atom emits its characteristic spectrum of light.

Let’s start to consider the planetary model for the simplest atom: the hydrogen atom.

Use Newton’s Second Law:

Fel = macircular , or ke2/r2 = mv2/r

(one equation, but two unknowns: v,r)

[Note that the theory should predict both v and r.]

We need more information, so try the law of Conservation of Energy:

E = KE + PE = (1/2)mv2 + -ke2/r = E

(a second equation, but introduce a third unknown, E; total unknowns: v, r, E)

Conservation of Angular Momentum:

L = mvr

(a third equation, but introduce a fourth unknown, L; unknowns: v, r, E and L.)

We have three equations and four unknowns.

Need some other piece of information or some other relation.

Bohr noted that Planck’s constant, h, had the units of angular momentum: L = mvr

(kg*m2/sec = Joule*sec)

so he tried this: L = nh

(quantize angular momentum).

L = n*(h/2n*

where  = h/2 (called h-bar)

This gave him four equations for four unknowns (treating the integer, n, as a known). From these he could get expressions for v, r, E and L.

In particular, he got:

r = n22/(meke2) = (5.3 x 10-11 m) * n2

(for n=1, this is just the right size radius for the atom) and

E = [-mek2e4/22]*(1/n2) = -13.6 eV / n2

(where 1 eV = 1.6 x 10-19 Joules).

This says the electron energy is QUANTIZED

In particular, when the electron changes its energy state (value of n), it can do so only from one allowed state (value of ninitial) to another allowed state (value of nfinal).

E = hf = [-13.6 eV]*[(1/ni2) - (1/nf2)] .

E = hf = [-13.6 eV]*[(1/nf2) - (1/ni2)]

In the case of ni = 3, and nf = 2,

E = (-13.6 eV)*(1/4 - 1/9) = 1.89 eV

E = hf = hc/ , so in this case,

emitted = hc/E =

(6.63x10-34 J-sec)*(3x108 m/s)/(1.89 x 1.6x10-19 J)

= 658 nm (red light).

Similarly, when ni = 4 and nf = 2, we get

E = 2.55 eV, andemitted = 488 nm

(blue-green); and

when ni = 5 and nf = 2, we get

E = 3.01 eV, andemitted = 413 nm

(violet).

ALL THREE MATCH THE ACTUAL SPECTRUM OF HYDROGEN!

This matching of theory with experiment is the reason Bohr made his assumption that

L = n(instead of L = nh).

• Note that we have quantized energy states for the orbiting electron.

• Note that for all nfinal = 1, we only get UV photons.

• Note that for all nfinal > 2, we only get IR photons.

Problems with the Bohr Theory:

• WHY is angular momentum quantized

(WHY does L=n need to be true.)

• What do we do with atoms that have more than one electron? (The Bohr theory does work for singly ionzed Helium, but what about normal Helium with 2 electrons?)

Problem with Bohr Theory: WHY L = n ?

• have integers with standing waves:

n(/2) = L

• consider circular path for standing wave:

n = 2r and so from Bohr theory:

L = mvr = n= nh/2get 2r = nh/mv = n

which means  = h/mv = h/p .

DeBroglie = h/mv = h/p

In this case, we are considering theelectron to be a WAVE, and the electron wave will “fit” around the orbit if the momentum is just right (as in the above relation). But this will happen only for specific cases - and those are the specific allowed orbits and energies that are allowed in the Bohr Theory!

The Introduction to Computer Homework on the Hydrogen Atom (Vol. 5, number 5) shows this electron wave fitting around the orbit for n=1 and n=2.

What we now have is a wave/particle duality for light (E&M vs photon), AND a wave/particle duality for electrons!

If the electron behaves as a wave, with

 = h/mv, then we should be able to test this wave behavior via interference and diffraction.

In fact, experiments show that electronsDO EXHIBIT INTERFERENCE when they go through multiple slits, just as the DeBroglie Hypothesis indicates.

Even neutrons have shown interference phenomena when they are diffracted from a crystal structure according to the DeBroglie Hypothesis:  = h/p .

Note that h is very small, so that normally  will also be very small (unless the mv is also very small). A small  means very little diffraction effects [1.22  = D sin()].

What we are now dealing with is the Quantum Theory:

• atoms are quantized (you can have 2 or 3, but not 2.5 atoms)

• light is quantized (you can have 2 or 3 photons, but not 2.5)

• in addition, we have quantum numbers

(L = n , where n is an integer)

There is a major problem with the wave/particle duality:

a) a wave with a definite frequency and wavelength (a nice sine wave) does not have a definite location. [At a definite location at a specific time the wave would have a definite phase, but the wave would not be said to be located there.]

[ a nice travelling sine wave = A sin(kx-t) ]

b) A particle does have a definite location at a specific time, but it does not have a frequency or wavelength.

c) Inbetween case: a group of sine waves can add together (via Fourier analysis) to give a semi-definite location: a result of Fourier analysis is this: the more the group shows up as a spike, the more waves it takes to make the group.

A rough drawing of a sample inbetween case, where the wave is somewhat localized, and made up of several frequencies.

A formal statement of this (from Fourier analysis) is: x * k

(where k = 2/, and  indicates the uncertainty in the value)

But from the DeBroglie Hypothesis, = h/p, this uncertainty relation becomes:

x * (2/) = x * (2p/h) = 1/2 , or

x * p = /2.

x * p = /2

The above is the BEST we can do, since there is always some experimental uncertainty. Thus the Heisenberg Uncertainty Principle says: x * p > /2 .

A similar relation from Fourier analysis for time and frequency:t *  = 1/2 leads to another part of the Uncertainty Principle (using E = hf):t * E > /2 .

There is a third part: * L > /2 (where L is the angular momentum value).

All of this is a direct result of the wave/particle duality of light and matter.

Let’s look at how this works in practice.

Consider trying to locate an electron somewhere in space. You might try to “see” the electron by hitting it with a photon. The next slide will show an idealized diagram, that is, it will show a diagram assuming a definite position for the electron.

We fire an incoming photon at the electron, have the photon hit and bounce, then trace the path of the outgoing photon back to see where the electron was.

incoming

photon

electron

screen

slit so we can

determine direction

of the outgoing

photon

outgoing

photon

electron

• Here the wave-particle duality creates a problem in determining where the electron was.

photon hits here

slit so we can

determine direction

of the outgoing

photon

electron

• If we make the slit narrower to better determine the direction of the photon (and hence the location of the electron, the wave nature of light will cause the light to be diffracted. This diffraction pattern will cause some uncertainty in where the photon actually came from, and hence some uncertainty in where the electron was .

We can reduce the diffraction angle if we reduce the wavelength (and hence increase the frequency and the energy of the photon).

But if we do increase the energy of the photon, the photon will hit the electron harder and make it move more from its location, which will increase the uncertainty in the momentum of the electron.

Thus, we can decrease the x of the electron only at the expense of increasing the uncertainty in p of the electron.

Let’s consider a second example: trying to locate an electron’s y position by making it go through a narrow slit: only electrons that make it through the narrow slit will have the y value determined within the uncertainty of the slit width.

But the more we narrow the slit (decrease y), the more the diffraction effects (wave aspect), and the more we are uncertain of the y motion (increase py) of the electron.

Let’s take a look at how much uncertainty there is: x * p > /2 .

Note that /2 is a very small number

(5.3 x 10-35 J-sec).

If we were to apply this to a steel ball of mass .002 kg +/- .00002 kg, rolling at a speed of 2 m/s +/- .02 m/s, the uncertainty in momentum would be 4 x 10-7 kg*m/s .

From the H.U.P, then, the best we could be sure of the position of the steel ball would be: x = 5.3 x 10-35 J*s / 4 x 10-7 kg*m/s

= 1.3 x 10-28 m !

As we have just demonstrated, the H.U.P. comes into play only when we are dealing with very small particles (like individual electrons or photons), not when we are dealing with normal size objects!

If we apply this principle to the electron going around the atom, then we know the electron is somewhere near the atom,

(x = 2r = 1 x 10-10 m)

then there should be at least some uncertainty in the momentum of the atom:

px > 5 x 10-35 J*s / 1 x 10-10 m = 5 x 10-25 m/s

Solving for p = mv from the Bohr theory

[KE + PE = Etotal, (1/2)mv2 - ke2/r = -13.6 eV

gives v = 2.2 x 106 m/s ] gives

p = (9.1 x 10-31 kg) * (2.2 x 106 m/s)

= 2 x 10-24 kg*m/s;

this means px is between -2 x 10-24 kg*m/s and 2 x 10-24 kg*m/s, with the minimum px being 5 x 10-25 kg*m/s, or 25% of p.

Thus the H.U.P. says that we cannot really know exactly where and how fast the electron is going around the atom at any particular time.

This is consistent with the idea that the electron is actually a wave as it moves around the electron.

But if an electron acts as a wave when it is moving, WHAT IS WAVING?

When light acts as a wave when it is moving, we have identified the ELECTROMAGNETIC FIELD as waving.

But try to recall: what is the electric field? Can we directly measure it?

Recall that by definition, E = F/q. We can only determine that a field exists by measuring an electric force! We have become so used to working with the electric and magnetic fields, that we tend to take their existence for granted. They certainly are a useful construct even if they don’t exist.

We have four LAWS governing the electric and magnetic fields: MAXWELL’S EQUATIONS. By combining these laws we can get a WAVE EQUATION for E&M fields, and from this wave equation we can get the speed of the E&M wave and even more (such as reflection coefficients, etc.).

But what do we do for electron waves?

What laws or new law can we find that will work to give us the wealth of predictive power that MAXWELL’S EQUATIONS have given us?

The way you get laws is try to explain something you already know about, and then see if you can generalize. A successful law will explain what you already know about, and predict things to look for that you may not know about. This is where the validity (or at least usefulness) of the law can be confirmed.

Schrodinger started with the idea of Conservation of Energy: KE + PE = Etotal .

He noted that

• KE = (1/2)mv2 = p2/2m, and that =h/p, so that p = h/= (h/2)*(2/) = k = p, so

KE = 2k2/2m

• Etotal = hf = (h/2)*(2f) = .

He then took a nice sine wave, and called whatever was waving, :

(x,t) = A sin(kx-t) = Aei(kx-t) .

He noted that both k and  were in the exponent, and could be gotten down by differentiating. So he tried operators:

(x,t) = A sin(kx-t) = Aei(kx-t) .

pop = i[d/dx] = i[-ikAe-i(kx-t)] = k

= (h/2)*(2/)* = (h/ = p .

similary:

Eop= i[d/dt] = i[-iAei(kx-t)] = 

= ((h/2)*(2f)* = (hf = E .

Conservation of Energy: KE + PE = Etotal

becomes with the momentum and energy operators:

-(2/2m)*(d2/dx2) + PE* = i(d/dt)

which is called SCHRODINGER’S EQUATION. If it works for more than the free electron, then we can call it a LAW.

What is waving here? 

What do we call ? the wavefunction

Schrodinger’s Equation allows us to solve for the wavefunction. The operators then allow us to find out information about the electron, such as its energy and its momentum.

To get a better handle on , let’s consider light: how did the E&M wave relate to the photon?

The photon was the basic unit of energy for the light. The energy in the wave depended on the field strength squared.

[Recall energy in capacitor, Energy = (1/2)CV2, where for parallel plates, Efield = V/d and

C// = KoA/d, so that

Energy = (1/2)*(KoA/d)*(Efieldd)2

= KoEfield2 * Vol, or Energy Efield2.]

Since Energy is proportional to field strength squared, AND energy is proportional to the number of photons, THEN that implies that the number of photons is proportional to the square of the field strength.

This then can be interpreted to mean that the square of the field strength is related to the probability of finding a photon.

In the same way, the square of the wavefunction is related to the probability of find the electron!

Since the wavefunction is a function of both x and t, the the probability of finding the electron is also a function of x and t!

Prob(x,t) = (x,t)2

Different situations for the electron, like being in the hydrogen atom, will show up in Schrodinger’s Equation in the PE part.

Different PE functions (like PE = -ke2/r for the hydrogen atom) will cause the solution to Schrodinger’s equation to be different, just like different PE functions in the normal Conservation of Energy will cause different speeds to result for the particles.